3
$\begingroup$

What I'm looking for is some sort of 'Bertini-theorem' for curves. Let $X$ be a Calabi-Yau manifold and let $C$ be a union of rational curves on $X$. Are there any techniques or results that would allow me to conclude that (perhaps some multiple of) $C$ can be deformed into an irreducible curve?

Edit: I'm looking for conditions on $C$ so that it moves. For example, if $C$ is a divisor on a K3, and $C^2>0$, then a general element in the linear system of some power is smooth by the usual Bertini.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ You probably want $C$ to be connected, otherwise there is little hope. Even so, I don't think this is possible in general: if $C$ is the union of two $(-2)$-curves on a K3 surface, the divisor $mC$ $(m\geq 1)$ never deforms. $\endgroup$
    – abx
    Commented Aug 8, 2014 at 14:47
  • $\begingroup$ This is a very annoying nitpick, but it isn't always true that a curve on a K3 with $C^2>0$ deforms to an irreducible one: $C$ could be of the form $F+A$ where $F$ is a $(-2)$-curve, $A^2 >>0$, and $A \cdot F=0$. (To get such examples, take $A$ to be the pullback of a sufficiently ample divisor on the contraction of $F$.) Then any member of any linear system $|mC|$ must contain $F$ as a component. Sorry I can't say anything helpful! $\endgroup$
    – user5117
    Commented Aug 8, 2014 at 15:20

1 Answer 1

Reset to default
3
$\begingroup$

If the curves are ample (that is, the normal bundles are $\oplus_i\mathcal{O}(k_i)$, with all $k_i>0$), they can be deformed to a smooth one (see Kollar, Rational Curves). Of course, this cannot happen on Calabi-Yau. However, if both curves sit inside a smooth part of rational connected variety $Z$ and are ample in $Z$, they have smooth deformation inside $Z$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .